1. 1 Homework Q & A Junior Navigation Chapter 7 The Celestial LOP

2. 2 Objectives: ■ Understand the altitude-intercept method of plotting and the relationships between Ho, Hc and intercept. ■ Identify the parts of the navigational triangle. ■ Compute altitude and azimuth of a celestial body using a scientific calculator. ■ Convert azimuth angle (Z) to azimuth (Zn)

3. 3 Practical Exercise 1. – 2. Follow the Student Manual for guidance.

4. 4 3. When the observer is closer to the GP than is the DR: a. Ho is greater than Hc. b. Hc is greater than Ho. c. intercept is away. d. the radius of the circle of position through the DR is less than that through the observer's actual position. Ref.: ¶ 16

5. 5 4. In completing a sight reduction of the Sun, your results indicate Ho = 35°17.4' and Hc = 35°36.2'. a. Find the value of the intercept (a). Ans: 18.8nm b. Is the intercept (a) toward (T) or away (A)? Ans: away, since Ho < Hc c. Are you closer to or further from the GP than is the DR? Ans: further from the GP Solution: Hc = 35°36.2'. Ho = 35°17.4' a = 18.8' = 18.8nm A Ref.: ¶ 14 - 18

6. 6 5. In completing a sight reduction of the Sun, your results indicate Ho = 43°45.3' and Hc = 43°38.8'. a.Find the value of the intercept (a). Ans: 6.5nm b. Is the intercept toward (T) or away (A)? Ans: toward, since Ho > Hc c. Are you closer to or further from the GP than is the DR? Solution: Hc = 43°38.8' Ho = 43°45.3' a = 6.5' = 6.5nm T Ref.: ¶ 14 - 18 Ans: closer to GP

7. 7 6. The elevated pole of the navigational triangle is: a. the pole having the same name as the body's declination. b. the pole having the same name as the DR latitude. c. always the North Pole. d. always the South Pole. Ref.: ¶ 21

8. 8 7. The distance from the elevated pole to the reference position or DR is: a. called co-latitude. b. called co-altitude. c. sometimes greater than 90. d. called latitude. Ref.: ¶ 24

9. 9 8. Declination is: a. one side of the navigational triangle. b. the angular distance from the observer to the GP of the body. c. always greater than 90°. d. the angular distance from the equator to the GP of the body. Ref.: ¶ 25, Fig. 7 – 5a

10. 10 9. In the navigational triangle, the distance from the DR to the GP of the body is: a. 90° - Dec. b. measured along a parallel of latitude. c. the radius of a celestial circle of position. d. measured along a meridian of longitude. Ref.: ¶ 26

11. 11 10. Azimuth (Zn) is: a. always an internal angle of the navigational triangle. b. always measured clockwise from true north. c. measured from either pole, depending on the hemisphere in which the observer is located. d. always less than 90°. Ref.: ¶ 28

12. 12 11. If the observer is in the southern hemisphere and the LHA of the sun is less than 180: a. Z is north and east. b. Z is south and east. c. Z is south and west. d. Z is north and west. Ref.: ¶ Figure 7-6c, Table 7-1

13. 13 12. If the observer is in the northern hemisphere and the sun is west of the observer: a. Zn = 360° - Z. b. Zn = 180° + Z. c. Zn = 180° - Z. d. Zn = Z. Ref.: ¶ Figure 7-6a, Table 7-2

14. 14 13. The two sides and angle used to solve the navigational triangle are: a. declination, azimuth angle, and altitude. b. co-latitude, co-altitude, and LHA. c. co-declination, co-altitude, and azimuth angle. d. co-latitude, co-declination, and LHA. Ref.: ¶ 21 – 29, Figure 7 -6

15. 15 Summary: LHA Hc Intercept Zn a. 316°41.9' 34°00.8' 3.5nm T 054° b. 58°43.4' 25°10.5' 10.8nm A 246° c. 39°18.1' 51°42.7' 5.6nm T 259° d. 322°03.3' 52°43.4' 14.9nm A 098° Ref.: ¶ 48 14. Using the values given below, obtain the intercept (a) and azimuth (Zn) by calculator solution. Click Button To View Note: Solutions use values for LHA, Lat, Dec, and Hc rounded to 5 decimal places and entered into the calculator as such. Values of arc sin and arc cos are left in the calculator at full precision and converted directly to Hc and Z. DR L DR Lo GHA Dec Ho a. 23°19.6'S 87°14.2'W 43°56.1' 13°17.2'N 34°04.3' b. 14°19.5'N 152°49.8'E 265°53.6' 14°26.8'S 24°59.7' c. 28°36.4'N 70°50.4'W 110°08.5' 15°58.5'N 51°48.3' d. 9°56.5'S 89°18.5'E 232°44.8' 12°40.5'S 52°28.5' Solution a. Solution b. Solution c. Solution d.

16. 16 Solution: Problem 14 a. LHA calculation: GHA 43°56.1´ corr +360° GHA 403°56.1´ Lo(W) – 87°14.2´ LHA 316°41.9´ Conversion of dm (deg, min) to decimal degrees LHA 316°41.9´ = 316.69833° Lat 23°19.6´S = 23.32667° Dec 13°17.2´N = -13.28667° Sin Hc = (cos 316.69833° × cos 23.32667° × cos –13.28667°) + (sin 23.32667° × sin -13.28667°) Hc = 34.01267° Hc = 34°00.8’ Cos Z = [sin –13.28667° – (sin 23.32667° × sin 34.01267)] ÷ (cos 23.32667 × cos 34.01267) Z = S 126.36580° E Z = S 126.4° E Zn = 180° – Z Zn = 53.6° rounded to 054° Hc = 34°00.8´ Ho = 34°04.3´ a = 3.5´ = 3.5nm T

17. 17 Solution: Problem 14 b. LHA calculation GHA 265°53.6´ Lo(E) +152°49.8´ LHA 418°43.4´ corr –360° LHA 58°43.4´ Conversion of dm to decimal degrees LHA 58°43.4´ = 58.72333° Lat 14°19.5´N = 14.32500° Dec 14°26.8´S = –14.44667° Sin Hc = (cos 58.72333° × cos 14.32500° × cos –14.44667°) + (sin 14.32500° × sin –14.44667°) Hc = 25.17579° Hc = 25°10.5´ Cos Z = [sin –14.44667° – (sin 14.32500° × sin 25.17579°) ÷ (cos 14.32500° × cos 25.17579°) Z = N 113. 86251° W Z = N 113.9° W Zn = 360° – Z = 360° – 113.9° Zn = 246.1° rounded to 246° Hc = 25°10.5´ Ho = 24°59.7´ a = 10.8´ = 10.8 nm A

18. 18 Solution: Problem 14 c. LHA calculation: GHA 110°08.5´ Lo(W) –70°50.4´ W LHA 39°18.1´ Conversion of dm to decimal degrees LHA 39°18.1´ = 39.30167° Lat 28°36.4´N = 28.60667° Dec 15°58.5´N = 15.97500° Sin Hc = (cos 39.30167° × cos 28.60667° × cos 15.97500°) + (sin 28.60667° × sin 15.97500°) Hc = 51.71109° Hc = 51°42.7´ Cos Z = [sin 15.97500° – (sin 28.60667° × sin 51.71109°)] ÷ (cos 28.60667° × cos 51.71109°) Z = N 100.65589° W = N 100.7° W Zn = 360° – Z = 360° – 100.7° Zn = 259.3° rounded to 259° Hc = 51°42.7´ Ho = 51°48.3´ a = 5.6´ = 5.6 nm T

19. 19 Solution: Problem 14 d. LHA calculation: GHA 232°44.8´ Lo(E) +89°18.5´ E LHA 322°03.3´ Conversion of dm to decimal degrees LHA 322°03.3´ = 322.05500° Lat 9°56.5´S = 9.94167° Dec 12°40.5´S = 12.67500° Sin Hc = (cos 322.05500° × cos 9.94167° × cos 12.67500°) + (sin 9.94167° × sin -12.67500°) Sin Hc = 0.79571 Hc = 52.72259° Hc = 52° 43.4´ Cos Z = [sin -12.67500° – (sin 9.94167° × sin 52.72259°)] ÷ (cos 9.94167°× cos 52.72259°) Cos Z = 0.13753 Z = S 82.09537°E = S 82.1°E Zn = 180° – Z = 180° – 82.1° Zn = 97.9° rounded to 098° Hc = 52° 43.4´ Ho = 52° 28.5´ a = 14.9´ = 14.9 nm A

20. 20 15. Refer to Chapter 6, Homework #6a & #6b. Complete the bottom portions of the USPS SR96 Form you started in those exercises to find the intercept (a) and azimuth (Zn) for those sights by the Law of Cosines method. For reference, the DR position given in those exercises and the answers you calculated for LHA, Dec, and Ho are provided below: DR L DR Lo LHA Dec Ho a. 30°06.8'N 85°43.6'W 51°12.8' 10°50.8'N 38°46.3' b. 41°18.0'N 73°06.8'W 323°31.7' 7°00.8'S 31°18.2' Click to View Solution 15 a. Summary: a. Hc = 38°48.0' a = 1.7nm A Zn = 259° b. Hc = 31°16.0' a = 2.2nm T Zn = 136° Click to View Solution 15 b. Ref.: ¶ 48

21. 21 Figure IM07 – 15a Solution for Problem 15 a. Given the data below find the intercept (a) and the azimuth (Zn) using the Law of Cosines Method. DR L DR Lo LHA Dec Ho 30°06.8'N 85°43.6'W 51°12.8' 10°50.8'N 38°46.3'

22. 22 Solution for Problem 15 b. Given the data below find the intercept (a) and the azimuth (Zn) using the Law of Cosines Method. DR L DR Lo LHA Dec Ho 41°18.0'N 73°06.8'W 323°31.7' 7°00.8'S 31°18.2' a. Figure IM07 – 15b

23. 23Q7 End Of Homework Q & A Junior Navigation Chapter 7 The Celestial LOP